Chemistry, 25.10.2019 08:43 mrmendrala

Acertain weak base has a kb of 8.50 × 10-7. what concentration of this base will produce a ph of 10.38?

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We are given a pH of 10.38
*pH + pOH = 14

10.38 + pOH = 14
----> pOH = 3.62

pOH = -log [OH]
----> [OH] = 2.3988x10^-4

Set up an I.C.E table using the concentration of OH found (y) and x being the unknown:
[base] + H2O ------> [conjugate acid] + OH
I x 0 0
C -y +y +y
E x-y y y

Given Kb, we can set up the equilibrium equation:
K = $\frac{prdoucts}{reactants}$
K = $\frac{y ^{2} }{x-y}$

We can substitute y for the concentration of OH we found, and K for the Kb value:
$8.50* 10^{-7} = \frac{(2.3988* 10^{-4})^{2} }{x-2.3988*10^{-4} }$

Solve the equation algebraically to get x.
X = 6.8*10^-2

Hope that helps.
If you need more clarification just let me know

Answer from: Ideen

Concentration of weak base : 0.0679

Further explanation

pH is the degree of acidity of a solution that depends on the concentration of H ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H ions of 10ⁿ.

According to Arrhenius, Base is a compound which in water releases OH-ions

Electrolytes (acids or bases) with a small degree of ionization are classified as weak electrolytes

OH concentration and molarity can be related to the formula

$\rm [OH^-] = \sqrt{Kb.M}$

Kb = base ionization constant

M = molarity

A certain weak base has a Kb of 8.50 × 10⁻⁷

pH = 10.38

pOH = 14 - 10.38

pOH = 3.62

$\rm [OH^-]=10^{-3.62}$

Then :

$\rm [10^{-3.62}^2] = \sqrt{8.5.10^{-7}.M}$

$\rm M=\dfrac{10^{-7.24}}{8.5.10^{-7}}\\\\M=1.18.10^{-1.24}\\\\M=\boxed{\bold{0.0679}}$

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Acertain weak base has a kb of 8.50 × 10-7. what concentration of this base will produce a ph of 10....

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